Solve the ODE with initial condition:

[joker2014]

y''-10y'+25=0

Solve the ODE with initial condition:

y(0) = 0,

y' (1) = 12e^5 .

I keep getting y=12/5e^5x when c1=0 and c2=12/5 ... but Answer key says y=2xe^5x

what am I doing wrong?

 

[axmls]

We have to see what you're doing to know what you're doing wrong.

 

[joker2014]

We have to see what you're doing to know what you're doing wrong.

I solved it and got general solution of y=c1e^5x+c2xe5x and the derivative is y'=c1(5e^5x)+c2(5xe^5x)
for y(0)=0 i found that c1=0
in y'(1)=12e^5 i found that c2=12/5

which then gives final solution of y=(12/5)e^5x

 

[axmls]

You need to use the product rule on the second term: $$(x e^{5x})' = x' e^{5x} + x (e^{5x})' = e^{5x} + 5x e^{5x}.$$

Also, as a side note, it's advised that you use the homework section for any difficulties you're having in homework next time you have a question.

 

[joker2014]

You need to use the product rule on the second term: $$(x e^{5x})' = x' e^{5x} + x (e^{5x})' = e^{5x} + 5x e^{5x}.$$

Also, as a side note, it's advised that you use the homework section for any difficulties you're having in homework next time you have a question.

Ohmygodd! You are right I did a silly mistake!

This isn't homework, only studying for exam. But i thinkyes better to ask there.

Thank you